Physics 211B : Solution Set #1
[1] Drude formula – Consider a hypothetical monovalent s-band metal with a simple cubic
crystal structure. The valence band dispersion is given by the tight binding result,
ε(k) = −2t cos(kx a) + cos(ky a) + cos(kz a) .
Compute the DC conductivity tensor σαβ . Show that σαβ = σ δαβ is diagonal, and obtain
an expression for σ. Numerically evaluate any integrals. The following result may prove
useful:
Zπ Zπ
p
du dv δ(cos u + cos v + 2λ) = 4 K( 1 − λ2 ) Θ(1 − λ2 ) ,
−π
−π
where K(x) is the complete elliptic integral of the second kind. Compare your result with
the Drude value you would obtain by approximating the band as parabolic, based on its
curvature at the zone center.
Solution: We must evaluate
2
Z
σαβ = 2e τ
d3k α β
v v
(2π)3
∂f 0
−
,
∂ε
Ω̂
where the dispersion is
ε(k) = −2t cos(kx a) + cos(ky a) + cos(kz a) .
The velocity is
v(k) =
1 ∂ε
2ta =
sin(kx a), sin(ky a), sin(kz a) .
~ ∂k
~
As a consequence of the cubic symmetry, σαβ = σ δαβ is diagonal, and we may write
2ta 2 Z d3k
∂f 0
2
σ = 2e τ
sin (kz a) −
.
~
(2π)3
∂ε
2
Ω̂
The monovalency condition means that there is one electron per unit cell, which in turn
means the chemical potential lies at µ = 0 so the s-band is half-filled. Changing variables
to u = kx a, etc.,
e2 τ t
σ= 3 2
2π ~ a
Zπ Zπ Zπ
du dv dw sin2 (w) δ(cos u + cos v + cos w) .
−π
−π
−π
1
We now use
Zπ Zπ
I(λ) = du dv δ(cos u + cos v + 2λ)
−π
−π
1−2|λ|
Z
=4
−1
(1)
√
1
1
q
2
2
1−x
1 − x + 2|λ|
(2)
Z1
ds
2
2
1 + |λ| − 2 1 + |λ|2 s2 + 1 − |λ| s4
−1
p
1 − |λ|
8
K
= 4 K 1 − λ2 .
=
1 + |λ|
1 + |λ|
= 4 1 − |λ|
q
(3)
(4)
Thus,
σ=
where
8C e2 τ t
·
.
π 3 ~2 a
(5)
Z1 p
q
C = dx 1 − x2 K
1 − 14 x2 .
0
Numerical integration gives C ' 2.59011.
Expanding the dispersion about the zone center, we find
ε(k) = −6t + ta2 k2 + O(k 4 ) ,
(6)
hence the effective mass is given by
ta2 ≡
~2
2m∗
=⇒
m∗ =
~2
.
2ta2
(7)
The Drude conductivity is then
σ=
ne2 τ
2ta2
e2 τ t
2
=
ne
τ
·
=
2x
·
,
m∗
~2
~2 a
(8)
where x ≡ na3 is the dimensionless density, equal to the average number of electrons per
unit cell (0 ≤ x ≤ 2). For a monovalent metal, the band is half filled, and x = 1, and the
prefactor 2x is 2. The exact value of the prefactor is 8C/π 3 ' 0.668.
[2] Thermal transport in a magnetic field – Consider a metal with a parabolic band ε(k) =
~2 k2 /2m∗ in the presence of a uniform magnetic field B. Use the Boltzmann equation*
to compute (a) the resistivity tensor ρ, (b) the thermal conductivity tensor κ, (c) the
thermopower tensor Q, and (d) the Peltier tensor u. Assume T is small, and work to
lowest nontrivial order in the temperature T . Also assume a constant relaxation time τ .
Does the Wiedemann-Franz law hold for the matrices κ and ρ ?
2
Solution: We begin with the Boltzmann equation,
h
i ∂f 0 e
ε−µ
∂δf
δf
−
=
v · eE +
∇T
v×B·
−
.
T
∂ε
~c
∂k
τ
We take ε(k) = ~2 k2 /2m∗ , and we write
δf (k) = k · A(ε)
⇒
vx
∂δf
~
∂δf
− vy
= ∗ (kx Ay − ky Ax ) ,
∂ky
∂kx
m
from which we obtain the linear relations

 

eEx +
1
+ωc τ 0
Ax
0
~τ
∂f
−ωc τ
eEy +
1
0 Ay  = ∗
m ∂ε
Az
eEz +
0
0
1
The solution is easily obtained:
 

Ax
1
0
∗
Ay  = ~τ /m ∂f +ωc τ
1 + ωc2 τ 2 ∂ε
Az
0
−ωc τ
1
0
ε−µ
T
ε−µ
T
ε−µ
T

∂x T
∂y T  .
∂z T

eEx +
0


eEy +
0
2
2
eEz +
1 + ωc τ
ε−µ
T
ε−µ
T
ε−µ
T

∂x T
∂y T  .
∂z T
The electrical and thermal currents are given by
Z
2e
jα = −
dε g(ε) ε Aα
3~
Z
2
jαq =
dε (ε − µ) g(ε) ε Aα .
3~
We now read off the transport coefficients from the relations
j = ρ−1 E − ρ−1 Q∇ T
jq = uρ−1 E − (κ + uρ−1 Q)∇ T .
We find

−ωc τ
0

ρ−1
1
0
2
2
0
1 + ωc τ


Z
1
−ωc τ
0
0
2e
ε(ε
−
µ)
∂f
τ
+ωc τ

ρ−1 Q = − ∗ dε g(ε)
−
1
0
3m
T
∂ε
1 + ωc2 τ 2
2
2
0
0
1 + ωc τ


Z
1
−ωc τ
0
0
2e
∂f
τ
+ωc τ

uρ−1 = − ∗ dε (ε − µ) g(ε) ε −
1
0
3m
∂ε
1 + ωc2 τ 2
0
0
1 + ωc2 τ 2


Z
1
−ωc τ
0
2
0
τ
2
ε (ε − µ)
∂f
+ωc τ

κ + uρ−1 Q =
dε g(ε)
−
1
0
∗
3m
T
∂ε
1 + ωc2 τ 2
2
2
0
0
1 + ωc τ .
2e
=
3m
Z

1
∂f 0
τ
+ωc τ
dε g(ε) ε −
∂ε
1 + ωc2 τ 2
0
3
We evaluate the integrals using the Sommerfeld expansion, and invoking the density of
states
√
2 (m∗ )3/2 √
ε.
g(ε) =
π 2 ~3
With an energy-independent relaxation time τ , we obtain


1
−ωc τ
0
2τ
ne
1
+ωc τ

%−1 = σ =
1
0
m∗ 1 + ωc2 τ 2
2
2
0
0
1 + ωc τ
The quantity ρ−1 Q is proportional to the same matrix as is ρ−1 , and one readily finds that
Q is a multiple of the unit matrix,
Q=−
π 2 kB2 T
2eεF
·I .
Here we have used the results
Z
∂f 0
= 21 π 2 (kB T )2 g(µ) + O(T 4 )
dε g(ε) ε (ε − µ) −
∂ε
g(εF ) =
We also find
u = T Qt (−B) = −
3n
m∗ kF
=
.
2
2
π ~
2εF
π 2 (kB T )2
Finally, the thermal conductivity tensor is

1
∗
nτ /m 
1 2 2
κ = 3 π kB T
+ωc τ
1 + ωc2 τ 2
0
2eεF
·I .
−ωc τ
1
0

0

0
2
2
1 + ωc τ .
Thus, the Wiedemann-Franz law,
κ=
π 2 2 −1
k Tρ
3e2 B
holds at the matrix level as well.
[3] Two bands – Calculate the frequency-dependent conductivity tensor for a direct gap
semiconductor in the presence of a magnetic field B = B ẑ. You should begin with the
Boltzmann equation in the relaxation time approximation (f 0 → f¯0 , δf → δf¯ for holes),
∂δf
∂f 0
e
∂δf
δf
− ev · E
−
v×B·
=− ,
∂t
∂ε
~c
∂k
τ
and the conduction and valence band dispersions,
εv (k) = εv0 − 12 ~2 mvαβ −1 k α k β
εc (k) = εc0 + 12 ~2 mcαβ −1 k α k β .
4
Assume the two bands behave independently, and solve the two Boltzmann equations for
the conduction electrons and valence holes. In each case, try a solution of the form
δf (k, t) = k µ Aµ (ε(k)) e−iωt .
The currents are
jc
jv
d3k
= 2e
(2π)3
Z
− vc δfc
+ vv δf¯v
.
Ω̂
Compute σαβ along principal axes of the effective mass tensors. You may assume that mv
and mc commute, i.e. they have the same eigenvectors. You should further assume that B
lies along a principal axis.
Solution: This problem is essentially solved in the notes in section 1.7. All that is left to
do is to sum the contributions from the valence and conduction bands:
σxx (ω) =
pv e2 τv
nc e2 τc
1 − iωτc
1 − iωτv
+
2
2 τ2
∗
∗
2
2
mc,x (1 − iωτc ) + ωc,⊥ τc
mv,x (1 − iωτv )2 + ωc,⊥
v
ωc,⊥ τc
ωv,⊥ τv
pv e2 τv
nc e2 τc
p
+
σxy (ω) = − p ∗
2
2 τ2
∗
∗
∗
2
2
mc,x mc,y (1 − iωτc ) + ωc,⊥ τc
mv,x mv,y (1 − iωτv )2 + ωc,⊥
v
σzz (ω) =
nc e2 τc
1
1
pv e2 τv
+
.
∗
∗
mc,z 1 − iωτc
mv,z 1 − iωτv
[4] Spin disorder resistivity (for the brave only!) – Consider an isolated trivalent Tb impurity ion in a crystal field. Application of Hund’s rules gives a total angular momentum
J = 6. A cubic crystal field splits this 13-fold degenerate multiplet into six levels: two
singlets, one doublet, and three triplets. The ground state is a singlet. Using the first Born
approximation, calculate the temperature-dependent resistivity in a free electron model
with a scattering Hamiltonian
Nimp
Himp = −A (g − 1)
X
δ(r − Rj ) S · Jj /~2 ,
j=1
where r and S are the conduction electron position and spin operators Rj and Jj are the
impurity position and angular momentum of the j th Tb impurity. A is the strength of the
exchange interaction, and g = 23 is the gyromagnetic factor.
(a) In general the relaxation time is energy-dependent: τ = τ (ε). Show that the resistivity
is given by ρ = m/ne2 hτ i, where the average is with respect to the weighting function
ε g(ε) (−∂f 0 /∂ε). Show also that
1
≤ hτ −1 i,
hτ i
which provides an upper bound for ρ which can often be computed.
5
(b) Use the results of (a) to derive the approximate expression for the resistivity ρ '
ρ0 pij Qji , where
(Ei − Ei )/kB T
e−Ei /kB T
pij = P −E /k T ·
k
B
1 − e−(Ei −Ej )/kB T
ke
2 2
2
Qij = 12 i J + j + 12 i J − j + i J z j ,
where the ionic energy levels are denoted by Ei and where the summations run over the
(2J + 1) crystal field states. Show that
ρ0 =
3πm (g − 1)2 A2 nimp
8e2 ~3 εF
.
(c) Show that the high temperature limiting value of ρ is J(J + 1) ρ0 . This is often called
the spin-disorder resistivity.
Solution: The collision term in Boltzmann equation is, from Fermi’s golden rule,
2
2π X X Ikσ [f ] =
Pi jkσ Himp ik0 σ 0 δ Ej + εk − Ei − εk0 fk0 σ0 (1 − fkσ )
~
ij k0 σ 0
2
2π X X 0 Pi f k σ Himp ikσ 0 δ Ej + εk0 − Ei − εk fkσ (1 − fk0 σ0 ) ,
−
~
0 0
ij k σ
where Pi is the Boltzmann weight for the ion in state i:
exp(−Ei /kB T )
.
Pi = P
n exp(−En /kB T )
Using plane wave states ψk (r) = V −1/2 exp(ik · r), the matrix element is obtained:
X
1
0
jkσ Himp ik0 σ 0 = − A (g − 1) ~−2
ei(k−k )·R jσ 0 S · J iσ .
V
R
We assume the impurity positions are uncorrelated, so that
2
2 X
jkσ Himp ik0 σ 0 2 = A (g − 1) jσ 0 S · J iσ 2
ei(k−k0 )·(R−R0 )
~2 V 2
0
R,R
=
Nimp
V2
A2 (g
−
~4
1)2
0 jσ S · J iσ 2 1 + (Nimp − 1)δkk0 .
The term proportional to δkk0 cancels when inserted into the collision integral. We are left
with
X Z d3k 0
2
0 2π 2
2
0 − Ei − εk jσ S · J iσ δ
E
+
ε
Ikσ [f ] = 5 A (g − 1) nimp
j
k
~
(2π)3
0
ijσ
Ω̂
n
o
× Pj fk0 σ0 (1 − fkσ ) − Pi fkσ (1 − fk0 σ0 ) .
6
0 is annihilated by I:
Note that the Fermi-Dirac distribution fkσ
e−βEj
1
eβ(εk −µ)
0
Pj fk00 σ0 (1 − fkσ
) = P −βEn β(ε 0 −µ)
δ Ej + εk0 − Ei − εk
β(ε
−µ)
+1
e k
+1 e k
ne
β(ε
−µ)
−βE
0
i
e k
1
e
δ Ej + εk0 − Ei − εk
= P −βEn β(ε 0 −µ)
β(ε
−µ)
+1
e k
+1 e k
ne
0
= Pi fkσ
(1 − fk00 σ0 ) .
We therefore write
0
fkσ = fkσ
+ δfkσ
0
fkσ
= fk0 =
,
1
eβ(εk −µ)
+1
and find
0
0
Pj (fk00 σ0 + δfk0 σ0 ) (1 − fkσ
− δfkσ ) − Pi (fkσ
+ δfkσ ) (1 − fk00 σ0 − δfk0 σ0 )
0
0
δfk0 σ0 − Pj fk00 σ0 + Pi 1 − fk00 σ0 δfkσ .
) + Pi fkσ
= Pj (1 − fkσ
The Boltzmann equation then takes the form
X ∂f 0
2π
jσ 0 S · J iσ 2
evk · E −
= 5 A2 (g − 1)2 nimp
∂ε
~
ijσ 0
Z 3 0
dk
0
0
δfk0 σ0
δ Ej + εk0 − Ei − εk
) + Pi fkσ
Pj (1 − fkσ
×
3
(2π)
Ω̂
− Pj fk00 σ0 + Pi (1 − fk00 σ0 ) δfkσ
.
We now sum both sides on σ and divide by two. Using
X X
jσ 0 S · J iσ 2 =
j J α i i J β j × σ0 S α σ σ S β σ0
σ
=
=
σ
αβ
1 2
3 ~ S(S + 1)δ
2
1 2
j J i 4~
α β j J i iJ j
,
we obtain
X 2 Z d3k 0
∂f 0
2π 2
2
j J i 0 − Ei − εk
evk · E −
= 5 A (g − 1) nimp
δ
E
+
ε
j
k
∂ε
~
(2π)3
ij
Ω̂
0
0
× Pj (1 − fkσ
) + Pi fkσ
δfk0 σ0 − Pj fk00 σ0 + Pi 1 − fk00 σ0 δfkσ
It is convenient to define the structure factor
X 2
S(u) ≡
Pi j J i δ(u − Ej + Ei ) .
ij
7
.
(9)
Note that S(−u) = e−βu S(u).
We assume a solution of the form
∂f 0
∂ε
and plug this into our Boltzmann equation. Further assuming an isotropic Fermi surface,
the δfk0 term integrates to zero since it is proportional to vk0 , aside from energy-dependent
(hence rotationally isotropic) factors. We then have
δfk = eτ (εk ) · E
1
π
= 3 A2 (g − 1)2 nimp
τ (εk )
2~
Z∞
Z 3 0
dk
du S(u)
δ(u + εk0 − εk )
(2π)3
−∞
Ω̂
h
i
× 1 − fk00 + e−βu fk0 .
Furthermore, since u = εk − εk0 , we have
1 − fk00 + e−βu fk0 =
1 − fk00
,
1 − fk0
giving
1
π
= 3 A2 (g − 1)2 nimp
τ (εk )
2~
Z∞
Z 3 0
1 − fk00
dk
0 − εk )
δ(u
+
ε
du S(u)
k
(2π)3
1 − fk0
−∞
Ω̂
(a) The conductivity is determined from
Z 3
∂f 0
dk
eτ
(ε
)
v
·
E
v
−
,
j = 2e
k
k
k
(2π)3
∂ε
Ω̂
which, for free electrons, gives
ne2 hτ i
m
,Z
Z
∂f 0
∂f 0
hτ i = dε ε g(ε) τ (ε) −
dε ε g(ε) −
.
∂ε
∂ε
σ=
Now the triangle inequality requires for any real functions f (ε) and g(ε)
Z
2 Z
Z
dε f (ε) h(ε) ≤
dε f (ε)
dε h(ε) ,
so taking
1/2
∂f 0
f (ε) ≡ ε g(ε) −
τ (ε)
∂ε
∂f 0
1 1/2
h(ε) ≡ ε g(ε) −
∂ε τ (ε)
8
we conclude
D 1 E
τ (ε) ·
≥1
τ (ε)
hence
ρ=
m
m D1E
≤
.
ne2 hτ i
ne2 τ
(b) Let us compute τ −1 . Noting that
∂f 0
= β f0 1 − f0 ,
−
∂ε
we have
D1E
τ
Z 3 Z 3 0
2m π 2
dk
dk
2
=
· 3 A (g − 1) βnimp
3
3n 2~
(2π)
(2π)3
Ω̂
Ω̂
Z
× du S(u) δ(u + εk0 − εk ) fk0 1 − fk00 vk2
Z
Z Z
πm A2 (g − 1)2
=
β
n
du
S(u)
dε dε0 δ(u + ε0 − ε)
imp
3~n (2π)6 ~4
Z
Z
0
0 0
× f (ε) 1 − f (ε ) dSε |v| dSε0 |v 0 |−1 .
The factor f 0 (ε) 1 − f 0 (ε − u) varies on a scale kB T . If kB T µ and if u µ then we can
approximate
Z
Z
2
dSε |v| dSε0 |v 0 |−1 ≈ 4πkF2 ,
and the energy integral becomes
Z
dε f 0 (ε) 1 − f 0 (ε) =
ω
.
eβω − 1
Thus,
m
=
A2 (g − 1)2 (kF /~)4 nimp
12π 3 ~n
Z
m D 1 E 3π mA2 (g − 1)2
nimp
ρ≤ 2
=
ne τ
8
e2 ~εF
Z
D1E
τ
and
du
du
βu
S(u) ,
−1
eβu
βu
S(u)
−1
eβu
since n = kF3 /3π 2 and εF = ~kF2 /2m. Note that
Z
du
X β(Ej − Ei )
βu
e−βEi 2
P
S(u)
=
j J i
eβu − 1
eβ(Ej −Ei ) − 1 n e−βEn
ij
X
≡
pij Qji ,
ij
9
where
β(Ej − Ei )
e−βEi
P −βE
n
eβ(Ej −Ei ) − 1 n e
2 1 + 2 1 i 2 z 2
Qji = j J i = 2 j J i + 2 j J i + j J i .
pij =
Therefore, the upper bound on the resistivity is ρ+ = ρ0 pij Qji , with
ρ0 =
3πm A2 (g − 1)2
nimp .
82
~3 εF
(c) At high temperatures, βu/(eβu − 1) → 1 and therefore pij → Pi , independent of j. In
this limit, then,
X 2 X
X
X
−βEi
j J i e−βEn
pij Qji =
e
ij
i
n
j
X X
=
i J2 i
e−βEn
n
i
= J(J + 1) .
Hence, ρ+ = J(J + 1)ρ0 at high temperatures.
[5] Cyclotron resonance in Si and Ge – Both Si and Ge are indirect gap semiconductors
with anisotropic conduction band minima and doubly degenerate valence band maxima. In
Si, the conduction band minima occur along the h100i (hΓXi) directions, and are six-fold
degenerate. The equal energy surfaces are cigar-shaped, and the effective mass along the
hΓXi principal axes (the ‘longitudinal’ effective mass) is m∗l ' 1.0 me , while the effective
mass in the plane perpendicular to this axis (the ‘transverse’ effective mass) is m∗t ' 0.20 me .
The valence band maximum occurs at the unique Γ point, and there are two isotropic hole
branches: a ‘heavy’ hole with m∗hh ' 0.49 me , and a ‘light’ hole with m∗lh ' 0.16 me .
In Ge, the conduction band minima occur at the fourfold degenerate L point (along the
eight h111i directions) with effective masses m∗l ' 1.6 me and m∗t ' 0.08 me . The valence
band maximum again occurs at the Γ point, where the hole masses are m∗hh ' 0.34 me and
m∗lh ' 0.044 me . Use the following figures to interpret the cyclotron resonance data shown
below. Verify whether the data corroborate the quoted values of the effective masses in Si
and Ge.
Solution: We found that σαβ = ne2 Γ−1
αβ , with
e
Γαβ ≡ (τ −1 − iω) mαβ ± αβγ B γ
c
 −1

∗
(τ − iω)mx
±eBz /c
∓eBy /c
(τ −1 − iω)m∗y
±eBx /c  .
=  ∓eBz /c
±eBy /c
∓eBx /c
(τ −1 − iω)m∗z
10
Figure 1: Constant energy surfaces near the conduction band minima in silicon. There are
six symmetry-related ellipsoidal pockets whose long axes run along the h100i directions.
The valence band maxima are isotropic in both cases, with
m∗hh (Si) ' 0.49 me
m∗hh (Ge) ' 0.34 me
m∗lh (Si) ' 0.16 me
m∗lh (Ge) ' 0.044 me .
With isotropic bands, the absorption is peaked at ω = ωc = eB/m∗ c, assuming ωc τ 1.
Writing ω = 2πf , the resonance occurs at a field
m∗ c
e
hc m∗
1
hf
=
·
·
·
2
e me 2πaB (e2 /aB )
m∗
= 3.58 × 10−7 G ·
· f [Hz]
me
m∗
= 8590 G ·
,
me
B(f ) = 2πf ·
11
Figure 2: Cyclotron resonance data in Si (G. Dresselhaus et al., Phys, Rev, 98, 368 (1955).)
The field lies in a (110) plane and makes an angle of 30◦ with the [001] axis.
where we have used
hc
= 4.137 × 10−7 G · cm2
e
~2
aB =
= 0.529 Å
me e2
h =4.136 × 10−15 eV · s
e2
= 27.2 eV = 2 Ry
aB
f = 2.40 × 1010 Hz .
Thus, we predict
Bhh (Si) ' 4210 G
Bhh (Ge) ' 2920 G
Blh (Si) ' 1370 G
Blh (Ge) ' 378 G .
All of these look pretty good.
12
Figure 3: Constant energy surfaces near the conduction band minima in germanium. There
are eight symmetry-related half-ellipsoids whose long axes run along the h111i directions,
and are centered on the midpoints of the hexagonal zone faces. With a suitable choice
of primitive cell in k-space, these can be represented as four ellipsoids, the half-ellipsoids
on opposite faces being joined together by translations through suitable reciprocal lattice
vectors.
Now let us review the situation with electrons near the conduction band minima:
Si : 6-fold degenerate minima along h100i
Ge : 4-fold degenerate minima along h111i (at L point)
m∗l (Si) ' 1.0 me
m∗l (Ge) ' 1.6 me
m∗t (Si) ' 0.20 me
m∗t (Ge) ' 0.08 me .
The resonance condition is that σαβ = ∞, which for τ > 0 occurs only at complex frequencies, i.e. for real frequencies there are no true divergences, only resonances. The location of
the resonance is determined by det Γ = 0. Taking the determinant, one finds
e2
m∗ e2
det Γ = (τ −1 − iω) m∗l · (τ −1 − iω)2 m∗t 2 + 2 Bz2 + ∗t 2 Bx2 + By2
.
c
ml c
13
Figure 4: Cyclotron resonance data in Ge (G. Dresselhaus et al., Phys, Rev, 98, 368 (1955).)
The field lies in a (110) plane and makes an angle of 60◦ with the [001] axis.
Assuming ωτ 1, the location of the resonance is given by
2
ω =
eBk
m∗t c
2
m∗
+ t∗
ml
eB⊥
m∗t c
2
,
where Bk ≡ Bz and B⊥ ≡ Bx x̂ + By ŷ. Let the polar angle of B be θ, so Bk = B cos θ and
B⊥ = B sin θ. We then have
2 n
o
m∗t
2
ω =
cos θ + ∗ sin θ
m
s l
m∗ m∗
t
B(f ) = 8600 G ·
cos2 θ + t∗ sin2 θ ,
me
ml
2
eB
m∗t c
2
where again we take f = ω/2π = 2.4 × 1010 Hz.
14
According to the diagrams, the field lies in the (110) plane, which means we can write
q
q
B̂ = 12 sin χ ê1 − 12 sin χ ê2 + cos χ ê3 ,
where χ is the angle B̂ makes with ê3 = [001].
Ge
We have
m∗t
= 0.051 ,
m∗l
m∗t
= 0.082
me
and we are told χ = 60◦ , so
B̂ =
q
3
8
ê1 −
q
3
8
ê2 + 12 ê3 .
The conduction band minima lie along h111i, which denotes a set of directions in real space:
±[111] : n̂ = ± √13 (ê1 + ê2 + ê3 )
⇒ cos2 θ = (B̂ · n̂)2 =
1
12
⇒ B = 1950 G
±[111̄] : n̂ = ± √13 (ê1 + ê2 − ê3 )
⇒ cos2 θ = (B̂ · n̂)2 =
1
12
⇒ B = 1950 G
±[1̄11] : n̂ = ± √13 (−ê1 + ê2 + ê3 ) ⇒ cos2 θ = (B̂ · n̂)2 =
⇒ cos2 θ = (B̂ · n̂)2 =
±[11̄1] : n̂ = ± √13 (ê1 − ê2 + ê3 )
√
7−2 6
12
√
7+2 6
12
⇒ B = 1510 G
⇒ B = 710 G .
All OK!
Si
Again, B lies in the (110) plane, this time with χ = 30◦ , so
B̂ =
q
1
8
ê1 −
q
1
8
ê2 +
q
3
4
ê3 .
The conduction band minima lie along h100i, so
±[001] : n̂ = ±ê3
⇒
cos2 θ = (B̂ · n̂)2 =
2
2
±[010] : n̂ = ±ê2
⇒
cos θ = (B̂ · n̂) =
±[100] : n̂ = ±ê1
⇒
cos2 θ = (B̂ · n̂)2 =
These also look pretty good.
15
3
4
1
8
1
8
⇒
B = 1820 G
⇒
B = 2980 G
⇒
B = 2980 G .